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Thread: Max/min on a continuous interval.

  1. #1
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    Max/min on a continuous interval.

    Suppose that $\displaystyle f(x)$ is continuous and $\displaystyle > 0$ on $\displaystyle I = \mathbb{R},$ and $\displaystyle \displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

    $\displaystyle (a)$ Prove $\displaystyle f(x)$ has no minimum on $\displaystyle I.$
    $\displaystyle (b)$ Prove $\displaystyle f(x)$ has a maximum on $\displaystyle I$ (note that $\displaystyle I$ is not compact).
    $\displaystyle (c)$ Prove $\displaystyle (b)$ under weaker hypothesis than positivity on all of $\displaystyle I$.
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  2. #2
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    Quote Originally Posted by cgiulz View Post
    Suppose that $\displaystyle f(x)$ is continuous and $\displaystyle > 0$ on $\displaystyle I = \mathbb{R},$ and $\displaystyle \displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

    $\displaystyle (a)$ Prove $\displaystyle f(x)$ has no minimum on $\displaystyle I.$
    $\displaystyle (b)$ Prove $\displaystyle f(x)$ has a maximum on $\displaystyle I$ (note that $\displaystyle I$ is not compact).
    $\displaystyle (c)$ Prove $\displaystyle (b)$ under weaker hypothesis than positivity on all of $\displaystyle I$.
    For (a) I'm assuming you mean a global minimum (otherwise this is not true) and it's obvious since that minimum would have to be $\displaystyle 0$ contradicting the fact that $\displaystyle f>0$.

    For (c) we're going to assume $\displaystyle f \geq 0$ for some $\displaystyle x \in \mathbb{R}$. Take an interval of the form $\displaystyle [-K,K]$ and take the maximum of f there say $\displaystyle f(x_0)=y_0>0$ (if it's zero we'll deal with it later). Now since $\displaystyle f(x) \rightarrow 0$ when $\displaystyle x \rightarrow \pm \infty$ we have two constants $\displaystyle K_1,K_2>0$ such that $\displaystyle \vert f(x) \vert < y_0$ whenever $\displaystyle x>K_1$ or $\displaystyle x<-K_2$ take $\displaystyle m= \max \{K_1,K_2 \}$ and so we have that $\displaystyle f$ attains a maximum $\displaystyle y$ in $\displaystyle [-m,m]$ with $\displaystyle y \geq y_0$ and since out of this interval the function never exceeds $\displaystyle y_0$, $\displaystyle y$ is a global maximum.

    If $\displaystyle y_0$ were $\displaystyle 0$ and $\displaystyle f \leq 0$ in I we're finished. If it's $\displaystyle f>0$ on an unbounded interval apply a similar argument to the above, if $\displaystyle f>0$ on a bounded interval it's clear it has a maximum. And this covers all cases.
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  3. #3
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    Can anyone show me where to start in order to prove the existence of a max by positivity?

    Thanks.

    [Forget it, I realized I can just pull it out of part (c)]
    Last edited by cgiulz; Oct 25th 2009 at 06:17 PM.
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